Prove that $$\sqrt{\frac{a^TA^2a}{a^TAa}}\le \sqrt{\|A\|}$$ where $A$ is a $n \times n$ symmetric positive definite matrix and $a \in \mathbb{R}^n \setminus \{0_n\}$.

The norm of a vector here is the Euclidean norm and norm of a matrix $A$ is $$\|A\|=\max\{\|Ax\|:\|x\|=1\}$$ and if $A$ is symmetric then $$\|A\|=\max\{|x^TAx|:\|x\|=1\}$$ and also equal to the largest eigenvalue of $A$.

We have from this min-max theorem that

$$\lambda_{\min}\|x\|^2\leq x^THx \le \lambda_{\max}\|x\|^2$$

where $\lambda_{\min},\lambda_{\max}$ are the smallest and largest eigenvalues of the Hermitian matrix $H$. So we have

$$\sqrt{\frac{a^TA^2a}{a^TAa}}\le \sqrt{\frac{\lambda^2_{\max}}{\lambda_{\min}}}$$

In order to prove the inequality in the question, we need to show that $\frac{\lambda^2_{max}}{\lambda_{\min}}\le \lambda_{\max}$ which implies $\frac{\lambda_{\max}}{\lambda_{\min}}\le 1$. We have $\|A\|=\lambda_{\max}$ and $\|A^{-1}\|=\frac{1}{\lambda_{\min}}$ but on the contrary it is a fact that $\|A\|\|A^{-1}\|\ge 1$. Where have I gone wrong and how do we prove this inequality?

  • 1
    $\begingroup$ i think your problem lies in the fact that you should not consider the maximal and minimal eigenvalues, because the vector a is the same in the denominator and the nominator. I'll write out the full solution in a sec $\endgroup$ Jan 25, 2018 at 11:22
  • $\begingroup$ So you're saying that I've overestimated by taking $\frac{\lambda^2_{max}}{\lambda_{min}}$ right? $\endgroup$
    – Picasso
    Jan 25, 2018 at 12:29
  • $\begingroup$ that is indeed what i'm saying $\endgroup$ Jan 25, 2018 at 12:53

2 Answers 2


Since $\rm A$ is symmetric and positive definite, it has a symmetric and positive definite square root $\mathrm A^\frac 12$ and $\lambda_{\max} \left( \mathrm A \right) = \| \mathrm A\|_2$, i.e., the spectral radius is equal to the spectral norm. Let $\rm v := \mathrm A^\frac 12 u$. Hence,

$$\dfrac{\mathrm u^\top \mathrm A^2 \mathrm u}{\mathrm u^\top \mathrm A \,\, \mathrm u} = \dfrac{\mathrm u^\top \mathrm A^\frac 12 \mathrm A \, \mathrm A^\frac 12 \mathrm u}{\mathrm u^\top \mathrm A^\frac 12 \mathrm A^\frac 12 \mathrm u} = \dfrac{\mathrm v^\top \mathrm A \, \mathrm v}{\mathrm v^\top \mathrm v} \leq \lambda_{\max} \left( \mathrm A \right) = \| \mathrm A\|_2$$

  • $\begingroup$ We actually need $\frac{u^TA^2u}{u^TAu} \leq \|A\|$. We know from here that there exists a symmetric positive definite square root of $A$, say which is $R$ here. Then we have $\frac{\|R^Tv\|^2}{\|v\|^2} \le \|R^T\|^2 = \|A\|$. $\endgroup$
    – Picasso
    Jan 26, 2018 at 11:08
  • $\begingroup$ I don't quite follow. We need $\sqrt{\|A\|}$ according to the problem, so in your proof we need to obtain $\|A\|$ as the upper bound. $\endgroup$
    – Picasso
    Jan 26, 2018 at 11:15
  • $\begingroup$ Yes, although the $\lambda$ part is unnecessary because we have for matrix $A$ and vector $v$, $\|Av\| \le \|A\|\|v\|$. So we directly have $\frac{\|A^{1/2}v\|^2}{\|v\|^2} \le \|A^{1/2}\|^2 = \|A\|$. $\endgroup$
    – Picasso
    Jan 26, 2018 at 11:36
  • $\begingroup$ I think it is now correct. The proof keeps getting shorter. $\endgroup$ Jan 26, 2018 at 11:48
  • $\begingroup$ We have $\frac{v^TAv}{v^Tv}\le \|A\|$ directly from the definition for norm of a symmetric matrix. Because $\frac{v^TAv}{v^Tv}=\frac{v^TAv}{\|v\|^2}=\frac{v^T}{\|v\|}A\frac{v}{\|v\|} \le \|A\|$. $\endgroup$
    – Picasso
    Jan 26, 2018 at 11:52

Let $A = X \Lambda X^{-1}$ be the eigenvalue decomposition for the matrix, this should always exist for a positive definite matrix. Also, $a = \mu_1 v_1 + ...$ denotes the decomposition of a into the eigenvectors of $A$.

Then $\sqrt{\frac{a^TA^2a}{a^TAa}} = \sqrt{\frac{\sum \mu_i^2 \lambda_i^2}{\sum \mu_i^2 \lambda_i}}$.

By calculating the derivative, we find that critical points allow only one $\mu_i$ to be non-zero. This proves that $a$ should be an eigenvector.

Should $a$ be an eigenvector with eigenvalue $\lambda_i$, $\sqrt{\frac{\sum \mu_i^2 \lambda_i^2}{\sum \mu_i^2 \lambda_i}}$ reduces to $\sqrt{\lambda_i}$ which is smaller than the norm of $A$.

  • $\begingroup$ Thanks. Is it possible to provide an answer that doesn't require the derivative? $\endgroup$
    – Picasso
    Jan 25, 2018 at 12:34
  • $\begingroup$ i can't think of something that wouldn't require a derivative. Perhaps some theory regarding rational functions. Not my area of expertise $\endgroup$ Jan 25, 2018 at 12:54

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